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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the sum of all two digit numbers which when divided by $4$ yields $1$ as remainder.

$\begin{array}{1 1}1196 \\ 1210 \\ 1312 \\ 1218 \end{array} $

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Toolbox:
  • $n^{th}$ term of an A.P.=$a+(n-1)d$
  • Sum of $n$ terms of an A.P=$\large\frac{n}{2}$$(t_n+a)$
Two digit numbers that yields remainder $1$ when divided by $4$ forms sequence
$13,17,21,........97$
This sequence is an A.P with first term $a=13$ common difference $d=4$ and
$t_n=97$
$\Rightarrow\:a+(n-1)d=97$
$\Rightarrow\:13+(n-1)4=97$
$\Rightarrow\:n-1=\large\frac{84}{4}$$=21$
or $n=22$
$\therefore$ Sum of the required numbers =$13+17+21+.......97$
$=\large\frac{n}{2}$$(t_n+a)=11(97+13)=11\times 110=1210$
answered Mar 29, 2014 by rvidyagovindarajan_1
 

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